The solution to this is ugly, right?

[Alec Mihailovs]

From: "Mitch Harris" <maharri at gmail.com>
Sent: Friday, May 30, 2008 1:34 PM

> On Fri, May 30, 2008 at 1:40 PM, Jack Brennen <jb at brennen.net> wrote:
>>  c = (a-b)/2
>>  u = (b-x)+c
>>
>> Solve:
>>
>>    (u+c)^(-1/2) + (u-c)^(-1/2) = 1 for u.
>>
>> The solution for u is still a quartic depending on c, but
>> it will at least appear a bit less ugly.
> 
> namely:
> -4 c^2 + c^4 +
> 4 c^2 u +
> (8 - 2 c^2) u^2 +
> -4 u^3 +
> u^4
> ==0

I think, it is wrong. I get 

u^4 - 4 u^3 - 2c^2 u^2 + 4c^2 u + c^4 + 4c^2 = 0

The solution can be expressed as

u = 1+3* 3F2(-1/2, 1/4, 3/4; 1/3, 2/3; -4 c^2/27)

where 3F2 is a hypergeometric function.

Alec Mihailovs